Question: A goblet contains $4$ red balls, $9$ green balls, and $8$ blue balls. If a ball is randomly chosen, what is the probability that it is not green?
Answer: There are $4 + 9 + 8 = 21$ balls in the goblet. There are $9$ green balls. That means $21 - 9 = 12$ are not green. The probability is $ \frac{12}{21} = \dfrac{4}{7}$.